Samacheer Kalvi Class 8 Maths Solutions Chapter 3 Algebra Exercise 3.10

 

Question 1. The sum of three numbers is 58. The second number is three times of two-fifth of the first number and the third number is 6 less than the first number. Find the three numbers.
Answer: Let the three numbers be \(a\), \(b\), and \(c\). We know that their sum is 58, so \( a + b + c = 58 \).
Let the first number be \(x\).
The second number \(b\) is three times two-fifth of the first number:
\( b = 3 \times \frac{2}{5} \times x = \frac{6}{5}x \)
The third number \(c\) is 6 less than the first number:
\( c = x - 6 \)
Now, substitute these expressions for \(b\) and \(c\) into the sum equation:
\( x + \frac{6}{5}x + (x - 6) = 58 \)
To remove the fraction, multiply the entire equation by 5:
\( 5x + 6x + 5(x - 6) = 58 \times 5 \)
\( 5x + 6x + 5x - 30 = 290 \)
Combine the \(x\) terms:
\( 16x - 30 = 290 \)
Add 30 to both sides:
\( 16x = 290 + 30 \)
\( 16x = 320 \)
Divide by 16 to find \(x\):
\( x = \frac{320}{16} \)
\( x = 20 \)
Now we can find all three numbers:
First number \( = x = 20 \)
Second number \( = \frac{6}{5}x = \frac{6}{5} \times 20 = 6 \times 4 = 24 \)
Third number \( = x - 6 = 20 - 6 = 14 \)
So, the three numbers are 20, 24, and 14. We can check by adding them: \( 20 + 24 + 14 = 58 \).
In simple words: We used algebra to find three unknown numbers. We set up equations based on the information given about their sum and how they relate to each other. By solving for 'x', we found the first number, then the others.

🎯 Exam Tip: Always define your variables clearly and write down the equations based on the problem statement. Double-check your calculations and substitute the values back into the original problem to verify your answer.

 

Question 2. In triangle ABC, the measure of ∠B is two-third of the measure of ∠A. The measure of ∠C is 20 more than the measure of ∠A. Find the measures of the three angles.
Answer: Let the measure of angle \(A\) be \(a^\circ\).
According to the problem, the measure of angle \(B\) is two-third of angle \(A\):
\( \angle B = \frac{2}{3} \times \angle A = \frac{2}{3}a \)
The measure of angle \(C\) is 20 more than angle \(A\):
\( \angle C = \angle A + 20 = a + 20 \)
We know that the sum of angles in any triangle is \(180^\circ\). So,
\( \angle A + \angle B + \angle C = 180^\circ \)
Substitute the expressions for \( \angle A, \angle B, \) and \( \angle C \):
\( a + \frac{2}{3}a + (a + 20) = 180 \)
Combine the \(a\) terms and move the constant to the other side:
\( a + \frac{2}{3}a + a = 180 - 20 \)
\( 2a + \frac{2}{3}a = 160 \)
To add the \(a\) terms, find a common denominator (3):
\( \frac{6a}{3} + \frac{2a}{3} = 160 \)
\( \frac{8a}{3} = 160 \)
Multiply both sides by 3:
\( 8a = 160 \times 3 \)
\( 8a = 480 \)
Divide by 8 to find \(a\):
\( a = \frac{480}{8} \)
\( a = 60 \)
Now, find the measures of all three angles:
\( \angle A = a^\circ = 60^\circ \)
\( \angle B = \frac{2}{3} \times 60^\circ = 2 \times 20^\circ = 40^\circ \)
\( \angle C = a^\circ + 20^\circ = 60^\circ + 20^\circ = 80^\circ \)
The three angles are \(60^\circ, 40^\circ,\) and \(80^\circ\). Their sum is \( 60 + 40 + 80 = 180^\circ \).
In simple words: We used the rule that angles in a triangle add up to 180 degrees. By writing each angle as an expression of 'a', we made an equation to find the value of 'a', then calculated each angle.

🎯 Exam Tip: Remember the fundamental properties of triangles, like the sum of interior angles being 180 degrees. This is key for solving angle-related problems.

 

Question 3. Two equal sides of an isosceles triangle are 5y - 2 and 4y + 9 units. The third side is 2y + 5 units. Find 'y' and the perimeter of the triangle.
Answer: An isosceles triangle has two equal sides.
Given the lengths of the two equal sides:
Side 1 \( = 5y - 2 \)
Side 2 \( = 4y + 9 \)
Since these sides are equal, we can set up an equation:
\( 5y - 2 = 4y + 9 \)
To solve for \(y\), move all \(y\) terms to one side and constants to the other (transposing):
\( 5y - 4y = 9 + 2 \)
\( y = 11 \)
Now that we have the value of \(y\), we can find the lengths of all three sides:
Length of the first equal side \( = 5y - 2 = 5(11) - 2 = 55 - 2 = 53 \) units
Length of the second equal side \( = 4y + 9 = 4(11) + 9 = 44 + 9 = 53 \) units
Length of the third side \( = 2y + 5 = 2(11) + 5 = 22 + 5 = 27 \) units
The perimeter of a triangle is the sum of the lengths of all three sides:
Perimeter \( P = 53 + 53 + 27 \)
Perimeter \( P = 133 \) units
So, the value of \(y\) is 11, and the perimeter of the triangle is 133 units.
In simple words: We used the property that an isosceles triangle has two equal sides to find the value of 'y'. Then we put 'y' back into the side lengths to calculate each side and finally added them up to get the perimeter.

🎯 Exam Tip: Remember the definition of an isosceles triangle (two equal sides) and how to calculate a perimeter (sum of all sides). Show substitution steps clearly.

 

Question 4. In the given figure, angle XOZ and angle ZOY form a linear pair. Find the value of x.
Answer: When two angles form a linear pair, their sum is always \(180^\circ\).
Given that \( \angle XOZ \) and \( \angle ZOY \) form a linear pair, we have:
\( \angle XOZ + \angle ZOY = 180^\circ \)
From the figure (and implied by the problem), let \( \angle XOZ = 3x - 2 \) and \( \angle ZOY = 5x + 6 \).
Substitute these expressions into the equation:
\( (3x - 2) + (5x + 6) = 180 \)
Combine the \(x\) terms and the constant terms:
\( (3x + 5x) + (-2 + 6) = 180 \)
\( 8x + 4 = 180 \)
Subtract 4 from both sides:
\( 8x = 180 - 4 \)
\( 8x = 176 \)
Divide by 8 to find the value of \(x\):
\( x = \frac{176}{8} \)
\( x = 22 \)
So, the value of \(x\) is \(22^\circ\).
We can also find the measures of the angles:
\( \angle XOZ = 3x - 2 = 3(22) - 2 = 66 - 2 = 64^\circ \)
\( \angle ZOY = 5x + 6 = 5(22) + 6 = 110 + 6 = 116^\circ \)
Check: \( 64^\circ + 116^\circ = 180^\circ \).
In simple words: A linear pair of angles always adds up to 180 degrees. We used this rule to create an equation with 'x' and then solved it to find what 'x' is.

🎯 Exam Tip: Understanding definitions like "linear pair" is essential. Make sure to combine like terms correctly when solving linear equations.

 

Question 5. Draw a graph for the following data:
Side of a square (cm)
Area (cm²)
Answer: Here is the data for the side of a square and its corresponding area:

To draw the graph, plot the points \((2, 4), (3, 9), (4, 16), (5, 25),\) and \( (6, 36) \) on a graph sheet with the side length on the x-axis and the area on the y-axis. Then, connect the plotted points with a smooth curve. When we plot this graph, we can see that the relationship between the side of a square and its area is not a straight line, which means it is not a linear relationship. The curve shows how the area grows faster as the side length increases. The graph will curve upwards, showing an exponential type of growth for the area.
In simple words: We take the given numbers for the side and area of a square and mark them as dots on a graph. Then we draw a curved line through these dots. This graph helps us see how the area of a square quickly gets bigger as its side gets longer.

🎯 Exam Tip: For plotting graphs, always label your axes clearly with units, choose an appropriate scale, and plot points accurately. Distinguish between linear and non-linear relationships by observing the shape of the graph (straight line vs. curve).

 

Question 6. Three consecutive integers, when taken in increasing order and multiplied by 2, 3 and 4 respectively, total up to 74. Find the three numbers.
Answer: Let the three consecutive integers in increasing order be \(x\), \(x + 1\), and \(x + 2\).
According to the problem, when these integers are multiplied by 2, 3, and 4 respectively, and then added together, the total is 74.
So, we can write the equation:
\( 2(x) + 3(x + 1) + 4(x + 2) = 74 \)
Now, expand and simplify the equation:
\( 2x + 3x + 3 + 4x + 8 = 74 \)
Combine the \(x\) terms and the constant terms:
\( (2x + 3x + 4x) + (3 + 8) = 74 \)
\( 9x + 11 = 74 \)
Subtract 11 from both sides:
\( 9x = 74 - 11 \)
\( 9x = 63 \)
Divide by 9 to find the value of \(x\):
\( x = \frac{63}{9} \)
\( x = 7 \)
Now, find the three consecutive integers:
First number \( = x = 7 \)
Second number \( = x + 1 = 7 + 1 = 8 \)
Third number \( = x + 2 = 7 + 2 = 9 \)
The three numbers are 7, 8, and 9. We can check the condition: \( 2(7) + 3(8) + 4(9) = 14 + 24 + 36 = 74 \). This matches the given total.
In simple words: We assumed the numbers are 'x', 'x+1', and 'x+2' since they are consecutive. We then built an equation from the problem's description, solved for 'x', and found all three numbers.

🎯 Exam Tip: When dealing with consecutive numbers, always represent them as \(x, x+1, x+2\), etc., for consecutive integers, or \(x, x+2, x+4\), etc., for consecutive even/odd integers.

 

Question 7. 331 students went on a field trip. Six buses were filled to capacity and 7 students had to travel in a van. How many students were there in each bus?
Answer: Let the number of students in each bus be 'x'.
There are 6 buses, so the total number of students in the buses is \(6 \times x = 6x\).
Apart from the students in the buses, 7 students traveled in a van.
The total number of students on the field trip is 331.
So, we can write an equation:
\( 6x + 7 = 331 \)
To solve for \(x\), first subtract 7 from both sides:
\( 6x = 331 - 7 \)
\( 6x = 324 \)
Now, divide by 6 to find the number of students in each bus:
\( x = \frac{324}{6} \)
\( x = 54 \)
Therefore, there were 54 students in each bus. This is a common way to distribute students when a few don't fit perfectly.
In simple words: We found out how many students were on the buses by taking the total students and subtracting those in the van. Then, we divided that number by the count of buses to see how many fit in each one.

🎯 Exam Tip: Clearly identify the knowns and unknowns. Formulate a linear equation to represent the problem, then solve for the variable using inverse operations.

 

Question 8. A mobile vendor has 22 items, some which are pencils and others are ball pens. On a particular day, he is able to sell the pencils and ball pens. Pencils are sold for Rs 15 each and ball pens are sold at Rs 20 each. If the total sale amount with the vendor is Rs 380, how many pencils did he sell?
Answer: Let 'p' be the number of pencils the vendor sold and 'b' be the number of ball pens he sold.
The total number of items is 22:
\( p + b = 22 \) (Equation 1)
Pencils are sold for Rs 15 each, and ball pens for Rs 20 each. The total sale amount is Rs 380:
\( 15p + 20b = 380 \) (Equation 2)
To solve this system of linear equations, we can use the elimination method.
First, divide Equation 2 by 5 to simplify it:
\( \frac{15p}{5} + \frac{20b}{5} = \frac{380}{5} \)
\( 3p + 4b = 76 \) (Simplified Equation 2')
Now, multiply Equation 1 by 3 so that the coefficient of 'p' matches in both equations:
\( 3 \times (p + b) = 3 \times 22 \)
\( 3p + 3b = 66 \) (Equation 3)
Next, subtract Equation 3 from Simplified Equation 2':
\( (3p + 4b) - (3p + 3b) = 76 - 66 \)
\( 3p - 3p + 4b - 3b = 10 \)
\( b = 10 \)
So, the vendor sold 10 ball pens. Now, substitute the value of \(b\) back into Equation 1 to find 'p':
\( p + 10 = 22 \)
\( p = 22 - 10 \)
\( p = 12 \)
The vendor sold 12 pencils. This shows how combining two pieces of information helps solve for both unknowns.
In simple words: We set up two math sentences for the number of items and the total money. By solving these two sentences together, we found out how many pencils were sold.

🎯 Exam Tip: For problems with two unknowns and two pieces of information, using simultaneous equations (like substitution or elimination) is the most efficient method. Be careful with signs during subtraction.

 

Question 9. Draw the graph of the lines y = x, y = 2x, y = 3x and y = 5x on the same graph sheet. Is there anything special that you find in these graphs?
Answer: To draw the graphs of these lines, we first find a few points for each equation by choosing values for \(x\) and calculating the corresponding \(y\) values:
(i) For the line \(y = x\):
When \(x = 1\), \(y = 1\)
When \(x = 2\), \(y = 2\)
When \(x = 3\), \(y = 3\)
(ii) For the line \(y = 2x\):
When \(x = 1\), \(y = 2(1) = 2\)
When \(x = 2\), \(y = 2(2) = 4\)
When \(x = 3\), \(y = 2(3) = 6\)
(iii) For the line \(y = 3x\):
When \(x = 1\), \(y = 3(1) = 3\)
When \(x = 2\), \(y = 3(2) = 6\)
When \(x = 3\), \(y = 3(3) = 9\)
(iv) For the line \(y = 5x\):
When \(x = 1\), \(y = 5(1) = 5\)
When \(x = 2\), \(y = 5(2) = 10\)
When \(x = 3\), \(y = 5(3) = 15\)
Now, plot these points for each equation on the same graph sheet. Draw a straight line through the points for each equation. All these lines will pass through the origin \((0,0)\).
**Observation:** When we plot these points and draw the lines, we notice that as the coefficient of \(x\) (which is the slope of the line) increases from 1 to 5, the lines become progressively steeper. This means the slope of the line keeps increasing. A higher coefficient of \(x\) makes the line rise faster for every step taken on the x-axis. For example, \(y=5x\) is much steeper than \(y=x\).
In simple words: We found some points for each math rule, then drew them as lines on one graph. We saw that lines with a bigger number next to 'x' went up much more quickly, meaning they were steeper.

🎯 Exam Tip: When graphing multiple lines on one sheet, label each line clearly with its equation. A common mistake is not choosing enough points to accurately draw the line, or incorrectly calculating the points. All lines in the form \(y=mx\) pass through the origin.

 

Question 10. Consider the number of angles of a convex polygon and the number sides of that polygon. Tabulate as follows: Use this to draw a graph illustrating the relationship between the number of angles and the number of sides of a polygon.
Answer: Here is the table showing the relationship between the number of sides and the number of angles of various convex polygons:

To draw the graph, we plot the number of sides on the x-axis and the number of angles on the y-axis. The data points will be \((3, 3), (4, 4), (5, 5), (6, 6), (7, 7),\) and \( (8, 8) \).
**Observation:** When these points are plotted and connected, they form a straight line passing through the origin (if we were to extend it). This indicates a direct and linear relationship: the number of angles in a convex polygon is always equal to its number of sides. This consistent relationship is why the points fall perfectly on a straight line.
In simple words: We listed different shapes and counted their sides and angles. Then, we drew a graph using these numbers. We saw that the number of sides is always the same as the number of angles, making a perfectly straight line on the graph.

🎯 Exam Tip: Recognize that for any polygon, the number of sides is always equal to the number of angles. When graphing such a relationship, ensure your axes are labeled correctly and the scale is consistent.